Question

The osmotic pressure of an unknown substance dissolved in 100 mL of water is measured at 298 K. The concentration of this substance is 25.5 kg m–3 and the osmotic pressure is 4.50 x 104 Pa. a) Determine the molecular weight of this substance. b) Calculate the change in the freezing point of water caused by the addition of this substance. (Keep in mind that the density of a dilute aqueous solution is approximately 1 g/mL).

Answer 1

Hi, the Finding of molecular wieght of unknow substance with given wieght can be done using VontHoff Osmatic pressure formula

=CST

= Osmatic pressure ( atm), C= concentration of Solute (Molarity), S= Universal gas constant (0.0820L. atm, K^-1, mol^-1), T= absolute temperature (K)

Now the give Osmatic pressure is 4.510⁴ Kpa ( i am assuming from the question)

Which = (1/101.325Kpa)4.510⁴ Kpa

= 444.115atm

Now, to find out molarity of substance

=CST, substitute the all the volues

444.115atm = C0.0820 L. atm, K^-1, mol^-1 298K

C = 444.115atm/(0.0820 L. atm, K^-1, mol^-1 298K)

C = 444.115/24.436 mol. L^-1

C = 18.147 mol. L^-1.

Now, give that substance dissolved in the 25.5 kg m^–3 concentration

The conversion is 1 kg/m³ = 0.001 g/cm³ So, now 0.0255gm per mL.

For 1000 mL = 25.5g in 1000 mL (1L)

Therefore, given wieght is 25.5g

Moles=Wt/Mw ( where M= Molarity, Wt= given wt, Mw= Molecular wt)

18.147 mol. L^-11L.= (25.5g/Mw )

Mw= 25.5/18.147

Mw= 1.4

Therefore the Molecular weight of the substance is 1.4amu.